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The image contains two mathematical problems where \(a\) and \(b\) represent the last two digits of a student identification number (NIM). In the example given, for a NIM ending in "61", \(a = 6\) and \(b = 1\).

### The two questions are:

1. **Find the solution set of** \(3x - 5 < |x + 3a + 2b|\).
   
2. **Find the solution set of** \(\left| 2a + b + \frac{2a + 3b}{2x + a + b} \right| > a + 2b\).

Now, let's break down and solve these questions using the example values of \(a = 6\) and \(b = 1\).

### Question 1: 
Solve \(3x - 5 < |x + 3a + 2b|\).

With \(a = 6\) and \(b = 1\), the equation becomes:
\[
3x - 5 < |x + 3(6) + 2(1)| = |x + 18 + 2| = |x + 20|
\]
We can break this inequality into two cases based on the absolute value:

- Case 1: \(x + 20 \geq 0\), so \(3x - 5 < x + 20\)
    \[
    3x - x < 20 + 5
    \]
    \[
    2x < 25 \quad \Rightarrow \quad x < \frac{25}{2} = 12.5
    \]
  
- Case 2: \(x + 20 < 0\), so \(3x - 5 < -(x + 20)\)
    \[
    3x - 5 < -x - 20
    \]
    \[
    3x + x < -20 + 5
    \]
    \[
    4x < -15 \quad \Rightarrow \quad x < -\frac{15}{4} = -3.75
    \]

Therefore, the solution to the inequality is \(x < -3.75\) or \(x < 12.5\), but \(x + 20 \geq 0\) only holds when \(x \geq -20\).

Thus, the final solution set is:
\[
x \in (-20, 12.5)
\]

### Question 2:
Solve \(\left| 2a + b + \frac{2a + 3b}{2x + a + b} \right| > a + 2b\).

Substitute \(a = 6\) and \(b = 1\) into the inequality:
\[
\left| 2(6) + 1 + \frac{2(6) + 3(1)}{2x + 6 + 1} \right| > 6 + 2(1)
\]
Simplify:
\[
\left| 12 + 1 + \frac{12 + 3}{2x + 7} \right| > 6 + 2 = 8
\]
\[
\left| 13 + \frac{15}{2x + 7} \right| > 8
\]

Now, we need to solve the inequality:
\[
\left| 13 + \frac{15}{2x + 7} \right| > 8
\]
This also splits into two cases:

- Case 1: \(13 + \frac{15}{2x + 7} > 8\)
\[
13 - 8 > -\frac{15}{2x + 7}
\]

Do you want to explore further solutions or have clarifications? 

---

Here are 5 follow-up questions for practice:

1. How does the absolute value affect the range of solutions?
2. Can you derive a general solution for inequalities involving absolute values?
3. What are the properties of rational inequalities?
4. How does adding constants inside absolute values impact the solution set?
5. What strategies help simplify complex fractions in inequalities?

Tip: Always check the domain of your variable in inequalities before concluding the solution set.

1. Tentukan himpunan penyelesaian dari 3x - 5 < |x + 3a + 2b|. 2. Tentukan himpunan penyelesaian dari |2a + b + (2a+3b)/(2x+a+b)| > a + 2b.

Learn how to solve complex algebraic inequalities, including absolute values and rational expressions. This problem explores key concepts in algebra and is suitable for students in Grades 10-12.

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